=lim(n tends to infinity) (cosx/2 * cosx/4 * ... * cosx/2n * sinx/2n)/sin (x/2n)
Obviously, judging from the double angle formula,
2sinacosa=sin2a
therefore
Cosx/2 n * sinx/2 n = [Sinks/2 (n-1]]/2
cosx /2^(n- 1)* sinx /2^(n- 1)=[sinx /2^(n-2)]/2
And so on,
(cosx/2 * cosx/4 *…*cosx/2^n *sinx/2^n)
=(sinx) / 2^n
therefore
Primitive restriction
=lim(n tends to infinity) [(sinx)/2 n]/sin (x/2 n)
=lim(n tends to infinity) (sinx)/[2 n * sin (x/2 n)]
=lim(n tends to infinity) [(sinx)/x]/[sin (x/2 n)/(x/2 n)]
Obviously, 2 n tends to infinity, that is, x/2 n tends to 0,
Then sin (x/2 n)/(x/2 n) tends to 1,
that is
Original limit = sinx /x